Find Number of Terms in Arithmetic Series Given Sum :
To find the first term, common difference, number of terms and sum of an arithmetic series, we use one of the formulas given below.
an = a + (n - 1)d
Sn = (n/2) [a + l] (or)
Sn = (n/2) [2a + (n - 1)d]
Question 1 :
Given a = 2 , d = 8, Sn = 90 find n and a n
Solution :
S n = (n/2) [2a + (n - 1) d]
90 = (n/2) [2 (2) + (n - 1) 8]
90 x 2 = n [4 + 8 n - 8]
180 = n [-4 + 8 n]
180 = -4 n + 8 n²
8 n² - 4 n - 180 = 0
Divide the whole equation by 4
2n² - n - 45 = 0
(n - 5) (2 n + 9) = 0
n = 5 or n = -9/2
The number of terms may not be negative or fraction.
a5 = a + 4 d
a5 = 2 + 4 (8)
a5 = 2 + 32
a5 = 34
Question 2 :
Given a = 8 , an = 62, Sn = 210 find n and d
Solution :
an = a + (n - 1) d
an = 8 + (n - 1) d
62 = 8 + (n - 1) d
62 - 8 = (n - 1) d
54 = (n - 1) d -----(1)
Sn = (n/2) [2a + (n - 1)d]
210 = (n/2) [ 2 (8) + (n - 1) d]
By applying the value of (n - 1) d in (1), we get
210 = (n/2)[16 + 54]
210 = (n/2) [70]
210/70 = (n/2)
3 x 2 = n
n = 6
(n-1) d = 54
(6-1)d = 54
5d = 54
d = 54/5
Question 3 :
Given an = 4, d = 2, Sn = -14 find n and a
Solution :
an = a + (n - 1) d
an = a + (n - 1) (2)
4 = a + 2 n - 2
4 + 2 = a + 2 n
6 = a + 2 n
a + 2 n = 6 ------(1)
an = l = 4
Sn = (n/2) [a + l]
-14 = (n/2) [ a + 4]
(-14 x 2) = n [ a + 4]
-28 = n[6- 2 n + 4]
-28 = n [10 - 2 n]
-28 = 10 n - 2 n²
2n² - 10 n - 28 = 0
n² - 5 n - 14 = 0
(n - 7) (n + 2) = 0
n = 7 or n = -2
By applying the value of n in (1), we get
a + 2(7) = 6
a + 14 = 6
a = 6 - 14
a = -8
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